Question

Find the value of y.
$y=\sqrt{log_2\:3.log_2\:12.log_2\:48.log_2\:192+16}-log_2\:12.log_2\:48+10$​

Answer 1

$\underline{\text{Solution :}}$

$\star \mathrm{log_{2}3=\frac{log_{e}3}{log_{e}2}=\frac{ln3}{ln2}}$

$\star \mathrm{log_{2}12=\frac{log_{e}12}{log_{e}2}=\frac{ln12}{ln2}=\frac{ln3+2ln2}{ln2}=\frac{ln3}{ln2}+2}$

$\star \mathrm{log_{2}48=\frac{log_{e}48}{log_{e}2}=\frac{ln48}{ln2}=\frac{ln3+4ln2}{ln2}=\frac{ln3}{ln2}+4}$

$\star \mathrm{log_{2}192=\frac{log_{e}192}{log_{e}2}=\frac{ln192}{ln2}=\frac{ln3+6ln2}{ln2}=\frac{ln3}{ln2}+6}$

$\mathrm{We\:consider\to \:\frac{ln3}{ln2}=x}$

$\boxed{\star}\mathrm{log_{2}3\times log_{2}12\times log_{2}48\times log_{2}192+16}$

$\mathrm{=\frac{ln3}{ln2}(\frac{ln3}{ln2}+2)(\frac{ln3}{ln2}+4)(\frac{ln3}{ln2}+6)+16}$

$\mathrm{=x(x+2)(x+4)(x+6)+16,\:since\:\frac{ln3}{ln2}=x}$

$\mathrm{=x(x+2)(x^{2}+10x+24)+16}$

$\mathrm{=x(x^{3}+12x^{2}+44x+48)+16}$

$\mathrm{=x^{4}+12x^{3}+44x^{2}+48x+16}$

$\mathrm{=x^{4}+36x^{2}+16+12x^{3}+8x^{2}+48x}$

$\mathrm{=(x^{2}+6x+4)^{2}}$

$\mathrm{Also,\:log_{2}12\times log_{2}48}$

$\mathrm{=\frac{log_{e}12}{log_{e}2}\times \frac{log_{e}48}{log_{e}2}}$

$\mathrm{=\frac{ln12}{ln2}\times \frac{ln48}{ln2}}$

$\mathrm{=\frac{ln3+2ln2}{ln2}\times \frac{ln3+4ln2}{ln2}}$

$\mathrm{=(\frac{ln3}{ln2}+2)(\frac{ln3}{ln2}+4)}$

$\mathrm{=(x+2)(x+4),\:since\:\frac{ln3}{ln2}=x}$

$\mathrm{=(x^{2}+6x+8)}$

$\therefore \mathrm{\sqrt{(log_{2}3\times log_{2}12\times log_{2}48\times log_{2}192)+16}-(log_{2}12\times log_{2}48)+10}$

$\mathrm{=\sqrt{(x^{2}+6x+4)^{2}}-(x^{2}+6x+8)+10}$

$\mathrm{=x^{2}+6x+4-x^{2}-6x-8+10}$

$\mathrm{=6}$

$\to \boxed{\mathrm{\sqrt{(log_{2}3\times log_{2}12\times log_{2}48\times log_{2}192)+16}-(log_{2}12\times log_{2}48)+10=6}}$

Answer 2

Answer:

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